PTA 10-排序6 Sort with Swap(0, i) (25分)

Posted 2020-09-27 gravitykey

题目地址

https://pta.patest.cn/pta/test/16/exam/4/question/678

 

5-16 Sort with Swap(0, i)   (25分)

Given any permutation of the numbers {0, 1, 2,..., N-1N?1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may apply the swap operations in the following way:

Swap(0, 1) => {4, 1, 2, 0, 3}
Swap(0, 3) => {4, 1, 2, 3, 0}
Swap(0, 4) => {0, 1, 2, 3, 4}

Now you are asked to find the minimum number of swaps need to sort the given permutation of the first NN nonnegative integers.

Input Specification:

Each input file contains one test case, which gives a positive NN (\le 10^5≤10?5??) followed by a permutation sequence of {0, 1, ..., N-1N?1}. All the numbers in a line are separated by a space.

Output Specification:

For each case, simply print in a line the minimum number of swaps need to sort the given permutation.

Sample Input:

10
3 5 7 2 6 4 9 0 8 1

Sample Output:

9

/*
评测结果
时间	结果	得分	题目	编译器	用时（ms）	内存（MB）	用户
2017-07-07 11:35	答案正确	25	5-16	g++	24	2	
测试点结果
测试点	结果	得分/满分	用时（ms）	内存（MB）
测试点1	答案正确	15/15	2	1
测试点2	答案正确	3/3	24	2
测试点3	答案正确	3/3	12	1
测试点4	答案正确	2/2	2	1
测试点5	答案正确	1/1	2	1
测试点6	答案正确	1/1	2	1

有点技巧性的题目，mooc上讲了，主要是找排序的环。有0环的交换次数为n-1，无零环为n+1
建了三个数组，A是存数，B是存i号元素放在了哪个位置，Checked是放这个元素有没有被访问过。

坑：必须加一个全局变量做搜索函数起点的备忘录用，否则搜索函数反复调用会导致超时
*/
#include<stdio.h>
#define MAXN 100000
int A[MAXN],B[MAXN],Checked[MAXN];
int gFindBeginPosition=0;
int FindUnchecked(int N)
{
	int i;
	for(i=gFindBeginPosition;i<N;i++)
	{
		if(!Checked[i])
			return gFindBeginPosition=i;
	}
	return -1;
}


int main()
{
	int i,N,p,sum;
	int ringCount=0;
	int nCount=0;
	int zeroInPosition=0;
	scanf("%d",&N);
	for(i=0;i<N;i++)
	{
		scanf("%d",&A[i]);
		B[A[i]]=i;
	}
	if(A[0]==0)
		zeroInPosition=0;
	else
		zeroInPosition=-2;
	
	while((p=FindUnchecked(N))+1)
	{
		Checked[p]=1;
		if(A[p]!=p)
			nCount++;
		else continue;
		while(!Checked[B[p]])
        {
			p=B[p];
			Checked[p]=1;
			nCount++;

		}
		ringCount++;
	}
	
	sum=nCount+ringCount+zeroInPosition;
	printf("%d",sum);
}